Solution:

Were you thinking of using the fastest person who takes one minute to travel to and fro till everyone has reached the other end? Yes that can be done but it will take 10 + 1 + 7 + 1 + 2 = 21 minutes in total.

If we pay attention towards finding a way to merge the one taking 7 minutes and the one taking 10 minutes, we will have to acknowledge the fact that one of them will also have to return back which will take much more time than 21 minutes.

What if we use the person taking 2 minutes to escort the one taking 1 minutes across? Let's come to the conclusion:

1 and 2 cross the bridge
2 returns with the torch
7 and 10 cross the bridge
1 comes back
1 and 2 cross the bridge

In such a manner, the total time taken will be
2 + 2 + 10 + 1 + 2 = 17 minutes.

Solution:

Even though the question might sound quite difficult, there is quite a simple solution for it if you light up your logics. All you have to do is turn two switches on and keep one off. Keep them in the same position for some 10 minutes and then shut one of them off. Hurry up into the room now.

There can be two possibilities - either the light are on or they are off. If they are on, you know which switch controls the lights as you have left only one switch on. If the lights are off, it can be either of the two remaining switches. All you have to do is touch the lights with your bare hand. If the lights are hot, it is the switch which you just turned off before entering the room and if the lights are cold, it is the switch which was left untouched and was off since the beginning.

Solution:

The argument of Chris Evans is quiet flawed as he fails to differentiate between purchasing and consuming.

Solution:

W02

F21: Friday the 21st.
S23: Sunday the 23rd.
T25: Tuesday the 25th.
T27: Thursday the 27th.
S29: Saturday the 29th.
M31: Monday the 31st.

Avoiding every other day, the next term must be 'Wednesday the 2nd' i.e. 'W02'

Solution:

24 times is the most common answer however that is completely wrong. The right answer will be 22 times.

Let us prove it by some simple mathematics.
Assume that it takes T hours for the minute hand to complete T laps. The hour hand will complete T/12 Laps in the same time.

Consider the situation when the hour hand and the minute hand will overlap for the first time; the minute hand would have completed one lap extra than the hour hands.
Thus, T =T/12 + 1

Considering the above expression we know that the first overlap will take place after t = 12/11 hours i.e. 1:05 am.

Similarly, the second overlap will take place when the minute hand would have completed two more laps than the hour hand.

Considering there are X laps,
T = T/12 + X

Everyone knows that a day comprises of 24 hours. Putting that in the equation we get
24 = 24/12 + X
Solve it and you will get
X = 22

Hence both of the hands will overlap 22 times in 24 hours.

Let us give you with those exact timings as well.
The hands will overlap at 12:00, 1:05, 2:10, 3:15, 4:20, 5:25, 6:30, 7:35, 8:40, 9:45, and 10:50. Do consider the fact that there will be no 11:55. It becomes 12:00.

Solution:

It is one of those problems which will keep you thinking for much time. But did you consider the fact that that the number of pennies are not told and the question asks if the stack will fit the room or not.

Also the size of the room is not mentioned and it is left to your intuition. Now consider the Empire State Building which is 102 stories tall or say 100 story tall on a rough note. Now that is hundred times more than an ordinary room measured on the inside. Now you will have to break the skyscraper-high column of pennies into about a hundred floor-to-ceiling-high columns. Now the question becomes if you can fit approximately hundred floor-to-ceiling penny columns in a room?

Surely! It will be just a ten by ten array of penny columns. Of course, every room will have enough space to set hundred pennies flat on the floor.

Solution:

The minimum number of drops required are eight.

We will begin from the floor 8. And then if it does not breaks, we will continue in the following fashion

8, (8+7), (8+7+6), (8+7+6+5), (8+7+6+5+4), (8+7+6+5+4+3), (8+7+6+5+4+3+2), (8+7+6+5+4+3+2+1)

= 8, 15, 21, 26, 30, 33, 35, 36

Suppose if the egg breaks from 20th floor
It will not break from eights
It will not break from fifteenth
It will break from twenty first
You will still have one egg remaining

Now start putting from sixteenth floor moving up one floor every time
It will not break till nineteenth and it will break of twentieth floor.

For the worst scenario possible, the maximum number of droppings will be eight only.

Solution:

Everyone will appoint any one person as a leader. Now everyone must follow some steps. Every prisoner will flip the first switch up at the first opportunity and also at the second opportunity. If the first switch is already up or they have flipped it up two times already, then they will flip the second switch. The leader however will have a different routine. He is the only one who will flip the switch down the first switch. Suppose if the switch is already down, he will flip the second switch. Also he will keep a count and when he has flipped the first switch down 44 times, he will declare that every prisoner has visited the room.

This is because once the leader has flipped down the switch 44 times, he will know that everyone has visited the room. If the switch was down initially, the other 22 prisoners will turn it up two times. If the first switch was already up, there will be one prisoner who will flip the switch up only once and the rest will do it twice.

Now this trick won't work when the leader has flipped the switch 23 times. Consider the fact that it is possible that 12 prisoners randomly and the leader have been taken to the switch room 24 times and the rest have not been taken till then. Remember that the first switch might have been up in the beginning itself. Thus the prisoners must flip the switch up twice. If they do it just once, the leader will never know whether to count till 22 or 23.

Solution:

Now this problem is a bit tricky. Before you start panicking about the solution let us give you a hint. You need to think in terms of binary numbers.

There are 1000 bottles which should be labeled from 1 to 1000 first. Now write those numbers in binary format.

Bottle Number 1 = 0000000001
Bottle number 2 = 0000000010
Bottle number 500 = 0111110100
Bottle number 1000 = 1111101000

Now the king will summon ten prisoners and label them from 1 to 10. The prisoner labeled as 1 will take a sip from those bottles that have a 1 in its least significant bit and the prisoner labeled as 10 will take a sip from every bottle that have a 1 in its most significant bit etcetera.

Let us frame an example. The wine from the bottle number 924 will be sipped by 10, 9, 8, 5, 4 and 3. Thus if that bottle will be poisoned, only those 6 prisoners will die.
After a month's time, all the king need to do is line up the prisoners in order and call each living prisoner as 0 and each dead prisoner as 1. N such a manner the number that the king will get will be of that bottle which is poisoned.

Now also note that if there would have been 1024 or more bottles, there would have been more prisoners required for the very same task.

Solution:

Divide the 9 gems into a group of three (randomly)

Now select any two group of gems and put them in a balance. If they are equal then, the heavier gem must be in the group you have left. If one of them is heavy, then that group contains the heavier gem.

In either case, take the group that contains the heavier gem and select any two gems among them (randomly). Weigh the two gems in the balance. If one is heavier, you know which one is the heavier gem of them all. If the weights are same, then the third gem is the heavier one in all.

In this manner, it will take you just two tries before you can determine the heavy gem out of the nine gems.